complex Lie groups, real Lie groups, and forgetful functors

Question

In the answer to this question What does it mean to be a real Lie group, there is some nice discussion.

My questions are:

1. I understand that the complex Lie algebra $\mathfrak{sl}(2,\mathbf{C})$ as the complexification of both of the real Lie algebra $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbf{R})$. I understand that the complex Lie group $SL(2,\mathbf{C})$ as the complexification of both of the $SU(2)$ and $SL(2,\mathbf{R})$. In this case, am I correct to see the two different real Lie groups (say $SU(2)$ and $SL(2,\mathbf{R})$) may have the same complexification to the same (isomorphic) complex Lie group (say $SL(2,\mathbf{C})$)?

Are the more than two (or three or more) different real Lie groups having the same complexification to the same (isomorphic) complex Lie group?

Reversely, am I correct to say that there are two (or more) different ways to apply the different forgetful functors to go from a complex Lie group to different real Lie groups?

2. About this:

"Do there exist two complex Lie groups which, after applying the forgetful functor, become isomorphic real Lie groups?"

What are some examples?

3. About this:

"Does there exist a real Lie group which is not isomorphic, as a real Lie group, to some complex Lie group with the forgetful functor applied?"

It is easy to get a real Lie group which as a smooth manifold that has an odd real dimension, such as $SO(2n+1)$ and $SU(2n+1)$. So I suppose this odd real dimension smooth manifold cannot be obtained from the complex Lie group (of a complex smooth manifold ) with the forgetful functor applied? Are there alternative possibilities than what I say?

Many thanks (in advance) for answering!

Answer 1

An incomplete answer, there is a lot to be said here.

For 1, real forms of complex Lie groups. Consider $GL(n,\mathbb{C})$. Then $GL(n, \mathbb{R})$ is a real form, and so is $U(n, \mathbb{C})$. But now we can take any hermitian form, of signature $(k, n-k)$ and consider the group $U(k, n-k)$. So there may be more than $2$ real forms.

For 2, we can consider complex tori, of form $\mathbb{C}/\Gamma$, where $\Gamma$ is a lattice ( this is related to elliptic curves, every complex elliptic curve is a torus. Now, as real Lie groups, any torus is iso to $\mathbb{R}^2 /\mathbb{Z}^2$. But in general, two tori are not iso even as complex varieties.

Still for $2$, I think if you stick to complex reductive groups then their structure as a real Lie group determines their structure as a complex Lie group.

For point 3, if a complex connected Lie group is compact, then it is abelian. Therefore, a compact real Lie group that is not abelian cannot have a complex structure.

But we do not have to be that restrictive. In general, a real Lie group does not come from a complex Lie group. One reason would be that in general, a real manifold does not come from a complex manifold ( does not have a complex structure). Also, a real Lie algebra in general does not have a complex structure.

$\bf{Added:}$ I think that the following holds: the real Lie group $G$ has a complex structure if and only if the Lie algebra $\mathfrak{g}$ has a complex structure.

Easy way to check that the real Lie algebra $\mathfrak{g}$ does not have a complex structure: $\dim_{\mathbb{R}}\mathfrak{g}$ is odd. Or: some subalgebra associated to $\mathfrak{g}$ ( like the center, or the derived subalgebra, etc) has an odd dimension. This shows right away that $gl(n,\mathbb{R})$ does not have a complex structure, because its center is $1$-dimensional.

What about if $\mathfrak{g}= sl(n, \mathbb{R})$, $n\ge 2$. This is a simple real Lie algebra. If it had a complex structure, then its complexification would not be simple. But its complexification is $sl(n, \mathbb{C})$, simple. Similarly we show that all the real forms of complex simple Lie algebras do not have a complex structure.

Answer 2

The answer by orangeskid is correct and makes good points on topic, but I would like to add another answer because I think there is a confusion / misunderstanding in the question which should be pointed out explicitly.

Namely, there are at least three ways to switch between complex and real Lie groups / algebras.

1. One way "up": Start with a real Lie group / algebra and complexify. If you start with something of real dimension $n$, then you get out something of complex dimension $n$. This is often called scalar extension. It can be described as a functor.

2. One out of two possible ways "down": Start with a complex Lie group / algebra and "forget" its complex structure, just "view it as a real Lie group / algebra". If you start with something of complex dimension $n$, then you get out something of real dimension $\color{red}{2}n$. This is often called scalar restriction. It can be described as a functor.

3. The other of two possible ways "down". Start with a complex Lie group / algebra and try (!) to find a (!) real Lie group / algebra such that, when you complexify it, you get the complex Lie group / algebra you started with. Such a thing is called a real form. If your complex object has complex dimension $n$, then a real form must have real dimension $n$.

The problem with 3 is that in general we neither have existence nor uniqueness, in other words A) not every complex Lie group / algebra has real forms, B) but some complex Lie groups / algebras have many different real forms.

Problem B) is what you ask about in your question 1, and as the other answer points out, indeed e.g. $SL_n(\mathbb C)$ has more than $\lfloor \frac{n}{2} \rfloor +1$ mutually non-isomorphic real forms, namely $SU(k,n-k)$ for $k = 0,..., \lfloor \frac{n}{2} \rfloor$ (actually there are even more: For $n \ge 3$, the standard split form $SL_n(\mathbb R)$ is not isomorphic to any of the list so far, and for $n \ge 4$ also $SL_n(\mathbb H)$ gives another form not isomorphic to the ones so far.) Likewise, all $SO(p,q)$ are real forms of $SO_{p+q}(\mathbb C)$ etc.

Problem A) has not been addressed so far here. A standard example of a complex Lie algebra which does not have a real form is described in the answer to Is every complex Lie algebra a complexification?.

Now problems A and B show quite clearly that the procedure no. 3, finding a real form to a complex Lie grouo / algebra, is not at all described by a functor (unless, maybe, if one restricted to tiny subcategories of the categories in question). In particular:

Do not mix up the forgetful functor "scalar restriction" with the task of "finding a real form"!

Because scalar restriction is a forgetful functor, but never finds you a real form (unless your object is $0$-dimensional), whereas real forms do not per se behave functorially, in fact they do not exist in general, and if they do, they are not unique in general.

More about all this in "Fact 2" of this answer https://math.stackexchange.com/a/4184237/96384, and in the bulk of https://math.stackexchange.com/a/3895802/96384 (for a long while this just talks about scalar extension and restriction and how they are not inverse to each other; see the "Added in response to comment" part to discuss real forms, i.e. method 3, born out of a desire to find "an inverse" to complexification which scalar restriction is not.)

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That all being said, I think you do mix up what you shouldn't in your questions 2 and 3. It turns out though that they can be answered either way you would clarify them:

If in question 2 you mean, are there non-isomorphic complex Lie groups whose scalar restrictions are isomorphic as real Lie groups: Yes, as pointed out in orangeskid's answer, all elliptic curves over $\mathbb C$, when one forgets their complex structure, just happen to be isomorphic to the real Lie group $\mathbb R^2 / \mathbb Z^2$. But there are many non-isomorphic (as complex Lie groups) elliptic curves over $\mathbb C$ (cf. Elliptic curves $\mathbb C/\Gamma , \mathbb C/\Gamma'$ are isomorphic iff $\Gamma=\lambda\Gamma'.$ or any introduction to elliptic curves).

If in question 2 you mean, are there non-isomorphic complex Lie groups which have real forms that are isomorphic: Then by definition, this cannot happen.

If in question 3 you mean, is there a real Lie group which is not the image of the scalar restriction functor applied to some complex Lie group: Then clearly, every real Lie group of odd dimension (like $\mathbb R$ itself) is an example. (Many even-dimensional ones exist as well, but there one needs to look closer.)

If in question 3 you mean, is there a real Lie group which is not a real form of a complex Lie group: Then again by definition, this cannot happen, because by definition any real Lie group is a real form of its own complexification.